Multiply (3x + 2) by (5x-7). So in this case, we are hitting two binomial terms, and I will explain to you Two equal ways to do this. The first is that you may have heard of it in the classroom and it is more mechanical Maybe it’s going to be faster, but we really don’t know what to do Then there is a method where you apply something you know It is a logical method. So first I will go through the preservation method that maybe where you will be going to use something called FOIL So let me write this, and you can see that when someone gives you A new method of conservation, as you are doing something mechanical. So FOIL came from First Outside – Let me write this way – FOIL since F came from First, and O came from Outside I came from Inside, and L came from Last. The reason I do not prefer these things It is that when you are 35 years old, you will not remember where FOIL came from So you won’t remember how to multiply the binomial statement, but let’s apply FOIL So first we will multiply the first expressions with each of the two-term expressions, that is, we multiply 3x by 5x 3x x 5x, the outside tells us to hit the external expressions, and in this case, we have 3x on the outside and we have – 7 on the outside. So this is 3x (-7). And inside Well, the internal expressions here are 2 and 5x. So 2 x 5x. Then finally we have the last two statements, which are 2 and -7 So the last two phrases are 2 x-7, 2 x-7. So what we do is make sure From that we multiply each phrase with every other phrase from here. What we do now is just a hit We use the distribution feature twice. Multiply by 3x (5x – 7). So 3x x (5x – 7) = (3x x 5x) + (3x – 7) And we multiply 2 by 5x-7 until we get these terms But anyway, let’s multiply this until we get the answer, 3x x 5x is 3 x 5 x x x x This equals 15x ^ 2. You can write that this x ^ 1 x x ^ 1 And multiply them until you get x ^ 2. And 3 x 5 = 15. And this phrase is 3 x -7 = -21 Then we have x here, followed by this expression which is 2 x 5 and equal to 10, x x = 10x And then finally we have this blue expression, 2 x -7 = -14, and we’re not done yet You can simplify it a little. We have two similar phrases here, we have these – let me find a new color – We have two phrases containing x ^ 1 or the phrase x here. So we have -21 of something and you added 10 Or in another way, we have 10 of something and you subtract 21 of them, so you get -11 from that We put the other expressions here, we have 15, 15x ^ 2 and then we have -14 so we are done Now I’ve said that I’ll show you another way to do this. I want to show you why The distribution feature brings us here without having to save FOIL So the distribution property tells us that if, look, if we were hitting something with a phrase, we would have to hit it With each phrase. So we can distribute, we can distribute 5x by 3 – Or actually we can, well, let me write that way– we can distribute 5x – 7 this all over 3x + 2 Let me change the arrangement, since we are used to the distribution from the left This is equivalent to (5x – 7) (3x + 2). I changed the places for the two phrases, and we can distribute all of this On each of these phrases. Now what would happen if I took (5x – 7) x 3x? Well, this Equivalent to 3x x (5x-7), I have distributed (5x – 7) x 3x and I will add 2 x (5x – 7) to it I distributed (5x – 7) over the 2 Now, we can use the distribution feature again, we can distribute 3x over 5x. We can distribute 3x by 5x We can distribute 3x over -7 We can distribute 2 by 5x, here We can divide the 2 by -7 Now if I do this then what will I get? 3x x 5x, this is here. If you multiply 3x x-7, that is this phrase here If you multiply 2 x 5x, this is that statement, and if you multiply 2 x -7, this represents this expression Here. So we got the same result that we got when using FOIL Now, FOIL could be faster if we wanted to use it and jump to this step I think it is important that you know that this is how it works Just in case you forgot this when you are 35 or 45 years old and you face the issue of hitting double-edged phrases You have to remember the distribution feature

Algebra